Trig Substitution Practice Problems with Answers

These trig substitution practice problems cover all three standard forms, a few direct formula cases, definite integrals and completing the square. Try each one first. The answers come first so you can check quickly, then the full solutions.

Pick the substitution from the form under the root:

Problems

Formula level

1. \displaystyle\int\frac{1}{\sqrt{4 - x^2}}\,dx
2. \displaystyle\int\frac{1}{9 + x^2}\,dx
3. \displaystyle\int\frac{1}{\sqrt{x^2 + 9}}\,dx

Full substitutions

4. \displaystyle\int\sqrt{1 - x^2}\,dx
5. \displaystyle\int\frac{x^2}{\sqrt{9 - x^2}}\,dx
6. \displaystyle\int\frac{1}{x^2\sqrt{4 - x^2}}\,dx
7. \displaystyle\int\frac{1}{(x^2 + 4)^{3/2}}\,dx
8. \displaystyle\int\frac{\sqrt{x^2 - 1}}{x}\,dx for x > 1

Definite integrals

9. \displaystyle\int_0^1\sqrt{1 - x^2}\,dx
10. \displaystyle\int_0^2\frac{1}{x^2 + 4}\,dx

Complete the square first

11. \displaystyle\int\frac{1}{x^2 + 2x + 5}\,dx
12. \displaystyle\int\frac{1}{\sqrt{3 - 2x - x^2}}\,dx

Answers

Step-by-step solutions

1. Let x = 2\sin\theta, so dx = 2\cos\theta\,d\theta and \sqrt{4 - x^2} = 2\cos\theta. The integral becomes \int d\theta = \theta:

2. Let x = 3\tan\theta, so dx = 3\sec^2\theta\,d\theta and 9 + x^2 = 9\sec^2\theta:

3. Let x = 3\tan\theta, so \sqrt{x^2 + 9} = 3\sec\theta and the integral becomes \int\sec\theta\,d\theta = \ln|\sec\theta + \tan\theta|. From the triangle, \sec\theta = \frac{\sqrt{x^2+9}}{3} and \tan\theta = \frac{x}{3}. The \ln 3 that comes out joins the constant:

4. Let x = \sin\theta, so the integral is \int\cos^2\theta\,d\theta = \frac{\theta}{2} + \frac{\sin\theta\cos\theta}{2}. With \sin\theta = x and \cos\theta = \sqrt{1 - x^2}, you get the answer in the table. See the integral of √(1 − x²).

5. Let x = 3\sin\theta. Then x^2 = 9\sin^2\theta, dx = 3\cos\theta\,d\theta and \sqrt{9 - x^2} = 3\cos\theta:

Since \sin\theta\cos\theta = \frac{x}{3}\cdot\frac{\sqrt{9 - x^2}}{3}, this gives \frac{9}{2}\arcsin\frac{x}{3} - \frac{x}{2}\sqrt{9 - x^2} + C.

6. Let x = 2\sin\theta. Then x^2 = 4\sin^2\theta and \sqrt{4 - x^2} = 2\cos\theta:

From the triangle, \cot\theta = \frac{\sqrt{4 - x^2}}{x}.

7. Let x = 2\tan\theta. Then (x^2 + 4)^{3/2} = 8\sec^3\theta and dx = 2\sec^2\theta\,d\theta:

From the triangle, \sin\theta = \frac{x}{\sqrt{x^2 + 4}}.

8. Let x = \sec\theta, so dx = \sec\theta\tan\theta\,d\theta and \sqrt{x^2 - 1} = \tan\theta:

With \tan\theta = \sqrt{x^2 - 1} and \theta = \operatorname{arcsec} x, you get the answer.

9. Use problem 4: \frac{1}{2}\left(\arcsin 1 + 0\right) - 0 = \frac{\pi}{4}. This is a quarter of the unit circle.

10. Use \frac{1}{2}\arctan\frac{x}{2}:

11. Complete the square: x^2 + 2x + 5 = (x + 1)^2 + 4. With u = x + 1, this is \int\frac{du}{u^2 + 4} = \frac{1}{2}\arctan\frac{u}{2}.

12. Complete the square: 3 - 2x - x^2 = 4 - (x + 1)^2. With u = x + 1, this is \int\frac{du}{\sqrt{4 - u^2}} = \arcsin\frac{u}{2}.

Common mistakes on these problems

• Choosing the wrong substitution. \sqrt{9 + x^2} needs \tan, not \sin. Match the sign pattern in the table above.
• Forgetting to replace dx. With x = 2\tan\theta, dx = 2\sec^2\theta\,d\theta. Leaving dx alone breaks every later step.
• Leaving the answer in \theta. An indefinite integral in x needs an answer in x. Use the triangle.
• Using trig substitution when u-substitution works. \int x\sqrt{4 - x^2}\,dx is one line with u = 4 - x^2.

Keep practicing

For unlimited problems with instant feedback, try the trig substitution lesson and the choosing a strategy lesson. For worked examples, read the trig substitution guide. For rational functions without roots, move on to partial fractions problems.